Fancy Showing Me How to Simplify This More?

(Fvt)²/c = E(Mv/t)vt/√( ½ Mc + E/M)

Since E((Fvt)²/c)=c²E(F²vt/c)=MpEc² where the expression E((Fvt)²/c) is also equivalent to E(Mv/t)v, then;

MpEc² – c²E(F²vt/c)=0

Divide M from both sides:

pEc² - c²E(F²vt/c/M)=0

Subtract Mv from both sides and substitute for p,

Ec² - c²E(F²vt/c/M)=-p [1]

Treating (F²vt/c/M) as the coefficient of c²E, then square it,

(F²vt/c/M)²

((F²)²v²t²/c²/M²)

Then add this to both sides of the equation [1],

Ec² - c²E(F²vt/c/M) + ((F²)²v²t²/c²/M²)=-p + ((F²)²v²t²/c²/M²)

Factor the left hand side:

(Ec²√(F²vt/c/M - 1) - F²vt/c/M)(Ec²√(F²vt/c/M - 1) - F²vt/c/M)

Which is (Ec²√(F²vt/c/M - 1) - F²vt/c/M)²

 

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More nonsense. \(p \neq mv\)

 

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You can't start from here if the units are inconsistent.

\([L^3 M^2 T^{-3}] =^{?} \frac{[L^4 M^2 T^{-4}]}{\sqrt{[L M T^{-1}] + [L^2 T^{-2}]}}\)
Clearly they are not only inconsistent, but they are the product of someone who has yet to achieve the least discipline in mathematics. Since your dimensions are inconsistent, if your statement were numerically true it would only be true in one particular measurement system and thus be numerology and not physics.

 

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It is numerically true. I have tested it seven times.

 

Slot in the correct values, and make up your own for mass and you will find these equations produce the correct units given.

 

That doesn't mean it's physically correct. Do 5 apples equal 5 oranges? The quantities are equal but that doesn't make an apple equal to an orange.

For instance, you wrote "½ Mc + E/M". You've added the quantity 0.5Mc to E/M. Well E/M has units of \([L]^{2}[T]^{-2}\) (since it's a velocity squared) and Mc has units of \([M][L][T]^{-1}\).

If someone asked you "How long till lunch?" would you reply "4 tons"? Or if you asked someone "How far away is London?" and they replied "50 degrees Centigrade" would you think they answered you correctly or would you point out that you don't measure distance in units of temperature.

This is a mistake you make time and time and time again. It's not even a complicated one! You only add quantities which have the same units. You don't add distance to time or subtract entropy from luminosity, you can only add and subtract the same physical quantities. 1 ton less than 4 tons is 3 tons. 30 seconds longer than a minute is 90 seconds. The 50 degrees C hotter than 40 degrees C is not 30 Newtons.

Now stop playing with your calculator and go find yourself a GCSE physics textbook and learn how to work with units!

 

Saxion is no longer with us, so this thread is no longer necessary.